I'm interested in the function $$f(v) = E\left[\big(\sum_i X_i v_i\big)^{2k}\right],$$ where $v$ is conditioned to $\|v\|_2=1$, $k\ge1$ is integer, and $X_i$ are iid random variables.
For which random distributions is $f$ maximized when all the weight of $v$ is concentrated on a single coordinate? That is, when $v=[1,0\dots,0]$.
Clearly when $X_i\sim\mathcal N(0,1)$ we get $E\left[\big(\sum_i X_i v_i\big)^{2k}\right]=C_k\|v\|^{2k}_2=C_k$ for some constant $C_k$ depending on $k$. Hence for the normal distribution the `layout' of $v$ doesn't matter. On the other hand, if $X_i$ is an $\alpha$-stable distribution, we'd have $E\left[\big(\sum_i X_i v_i\big)^{2k}\right]=C_k\|v\|^{2k}_\alpha$, which is clearly maximized at $v=[1,0\dots,0]$ when $\alpha<2$.
It seems to me that $f(v)$ will be convex for any random distribution that has Gaussian moments or larger. I have checked it for the exponential distribution and chi-squared as well, but I don't know how to prove this in general.
Is $f(v)$ convex for any distribution with $E\left[X^{2k}\right]\ge (2k-1)!!$ (rhs being the $2k$th moment of a standard normal distribution.)
In fact, we can show that $$ g(v) = E\left[(\sum_{i} v_i X_i)^{2k}\right]^{\frac{1}{2k}}$$ is subadditive, homogeneous of degree $1$ over $v\in\mathbb{R}^n$ (this implies convexity.) Subadditivity comes from Minkowski's inequality which is implying that $$\begin{eqnarray} g(v+w) &=& E\left[(\sum_{i} v_i X_i +\sum_{i} w_i X_i)^{2k}\right]^{\frac{1}{2k}}\\&\leq& E\left[(\sum_{i} v_i X_i)^{2k}\right]^{\frac{1}{2k}} + E\left[(\sum_{i} w_i X_i)^{2k}\right]^{\frac{1}{2k}}\\&=&g(v) + g(w). \end{eqnarray}$$ Homogeneity can be shown easily: $$ g(tv) = E\left[(\sum_{i} tv_i X_i)^{2k}\right]^{\frac{1}{2k}}=t E\left[(\sum_{i} v_i X_i)^{2k}\right]^{\frac{1}{2k}} = tg(v),\quad\forall t\geq 0. $$ This proves $$ g(tv + (1-t)w) \leq g(tv) + g((1-t)w) =tg(v) + (1-t)g(w) ,$$ for all $t\in [0,1]$ and $v,w\in\mathbb{R}^n$. Since $g$ is a non-negative convex function, and the map $t\in [0,\infty)\mapsto t^{2k}$ is non-decreasing convex, we have $f(v) = g(v)^{2k}$ is also convex.